The Shell Method

Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the x-axis.
$$y=\frac 1 x$$

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2. Relevant equations

$$Volume=2\pi\int_a^b p(y)h(y)dy$$

3. The attempt at a solution

I see that there are two shells, therefore, I would do the integral twice. Meaning, ##Volume=2\pi\int_a^b p(y)h(y)dy+2\pi\int_a^b p(y)h(y)dy##

I'm having trouble identifying what h(x) are for both integrals. I know that you can change ##y=\frac 1 x## to be ##x=\frac 1 y## and I know that for both p(y)=y. I think the dashed lines indicates that ##y=\frac 1 2## is the axis in which it is being rotated.
How do you determine h(x)? Once I understand how to determine h(x) for both integrals, I know how to solve the rest of the problem.

I see that there are two shells, therefore, I would do the integral twice. Meaning, ##Volume=2\pi\int_a^b p(y)h(y)dy+2\pi\int_a^b p(y)h(y)dy##

You won't do the integral twice -- you need to evaluate two different integrals with different limits of integration. Also, it would help you to have more meaningful expressions than p(y) and h(y). For each integral the typical volume element will be ##2\pi \cdot \text{radius} \cdot \text{width} \cdot dy##
In both integrals the radius will be y (i.e., the y-coordinate on the horizontal strip being revolved), but the width of the element that is being revolved is different in the two integrals.

In the lower region, the width is very simple. In the upper region, the width varies, but it will always be the x-value on the curve (expressed in terms of y) minus the x-value on the vertical line. Since you know the equation of the curve, you can write the x-value on the curve in terms of the y-coordinate there. You're on the right track below.

Jovy said:

I'm having trouble identifying what h(x) are for both integrals. I know that you can change ##y=\frac 1 x## to be ##x=\frac 1 y## and I know that for both p(y)=y. I think the dashed lines indicates that ##y=\frac 1 2## is the axis in which it is being rotated.
How do you determine h(x)? Once I understand how to determine h(x) for both integrals, I know how to solve the rest of the problem.

You won't do the integral twice -- you need to evaluate two different integrals with different limits of integration. Also, it would help you to have more meaningful expressions than p(y) and h(y). For each integral the typical volume element will be ##2\pi \cdot \text{radius} \cdot \text{width} \cdot dy##
In both integrals the radius will be y (i.e., the y-coordinate on the horizontal strip being revolved), but the width of the element that is being revolved is different in the two integrals.

In the lower region, the width is very simple. In the upper region, the width varies, but it will always be the x-value on the curve (expressed in terms of y) minus the x-value on the vertical line. Since you know the equation of the curve, you can write the x-value on the curve in terms of the y-coordinate there. You're on the right track below.

the x-value on the curve is ##\frac 1 y## but what is the x-value on the vertical line? Do you mean what x is in ##y=\frac 1 x##?